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I intuitively understand the gradient in a mathematical sense, especially the fact that it points in the direction of maximum increase and easily tells us the function's sensitivity to change in each direction.

However, in physics it seems more complicated, due to the existence of units. For example, when differentiating a single-variable scalar function with respect to, say, $x$, you know that the units will be divided by those of $x$ (i.e. the derivative of position ($\mathrm{m}$) with respect to time ($\mathrm{s}$) is in $\mathrm{m/s}$). However, gradients don't specify a single variable, but rather measure the sensitivity to change of every variable.

So for example, if we had a function $f(x,y,z,t)$ which describes the heat you feel from an explosion, depending on distance and time, what would be the units of the gradient? Alternatively, if we describe the speed of a particle in water as a function of distance and time, what would the gradient be?

I'm asking this because in cases where I'm given an equation like $$E = - \nabla V$$

I want to know if I can immediately infer the units of $E$ by knowing the units of $V$, or vice versa (not this specific E&M case, but in general).

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    $\begingroup$ Typically the units of gradients are specified if they're not derivatives of position. If not, it's usually safe to assume that the derivatives in the gradient are spatial. $\endgroup$ – probably_someone Feb 19 '18 at 21:43
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Yes. A gradient has dimensions of one over whatever you're differentiating by. So for a spatial gradient like that, it has dimensions of $1\over\rm distance $, or $1\over\rm m $ in SI.

Which gives you the units of electric field, $\rm V\over m $.

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Think even simpler and just consider a 1-D derivative for now.

Say that $x$ has the units of distance, and $f(x)$ has some units..

The quantity $\frac{d f}{dx}$ will have the units of $f(x)$ divided by distance. So if $f$ has the units of distance, $\frac{d f}{dx}$ will have the units of distance / distance, which is unitless.

Here is a simple example: say $f(x) = 2x$. Then $f(x)$ clearly has the units of distance because you are just multiplying a distance by $2$. But $\frac{d f}{dx} = 2$ is unitless.

On way to remember this is to note that, just by looking at $\frac{d f}{dx}$, we can see that a distance $dx$ is in the denominator.

$$\frac{d f}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$

The numerator has whatever units $f(x)$ has, and the denominator has the units of distance.

Here's one last way to see that $\frac{d f}{dx}$ has the units of $f(x)$ divided by distance. Take any distance scale, say a meter. Then we can express $x$ by a dimensionless number (let's call it $r$) times 1 meter.

$$x = r \times \text{1 meter}$$

$r$ is just $x$ measured in meters.

We then see

$$\frac{d f}{dx} = \frac{d f}{d(r \times \text{1 meter})} = \frac{1}{\text{1 meter}} \frac{d f}{dr}$$

(This is just a slightly slicker way to write the chain rule of differentiation.) $\frac{d f}{dr}$ has the sames units as $f(x)$ because $r$ is dimensionless.

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